Let's say we have a vector operator $ \hat{\mathbf{A}} $ whose components are operators $ \hat A_1, \hat A_2, \hat A_3, … $. How do you do the commutator between two vector operators. Is it correct to say
$$ [\hat{\mathbf{A}},\hat{\mathbf{B}}]= \hat{\mathbf{A}}\cdot\hat{\mathbf{B}}-\hat{\mathbf{B}}\cdot \hat{\mathbf{A}} $$
If so, what does this dot product mean and how do I operate? Since the elements are operators they can be thought as matrix, does that mean that the components should be multiplied as so and then add up? Is there any other type of product defined for this commutator?
Best Answer
It's non-standard notation. In your notation $\mathbf{A}$ is a vector with components $A_i$. Passing to a different frame/coordinate system, the components would transform as $$A_i \mapsto R_{ij} A_j$$ where $R_{ij}$ is matrix describing the coordinate change. The most general commutator is therefore a matrix: $$[A_i,B_j] = A_i B_j - B_j A_i$$ and it transforms as you'd expect (i.e. as a rank-2 tensor). The definition you're giving is a scalar one, namely the contraction with the $\delta_{ij}$ tensor: $$\delta^{ij} [A_i,B_j] = A_i B_i - B_i A_i = \mathbf{A} \cdot \mathbf{B} - \mathbf{B} \cdot \mathbf{A}.$$ This is similar to the usual grad/div stuff. Namely suppose that you have a vector field $\mathbf{A}(\mathbf{x})$ with components $A_i(\mathbf{x})$, and $\mathbf{x}$ itself is a vector $\mathbf{x} = x_j$. Then the most general tensor ("grad") you can build is $$\nabla \mathbf{A}(\mathbf{x})$$ with components $$\nabla_i A_j(\mathbf{x}) = \frac{\partial}{\partial x^i} A_j(\mathbf{x}).$$ Of course, you can take the trace of that tensor to get the "div": $$\nabla \cdot \mathbf{A}(\mathbf{x}) = \delta^{ij} \nabla_i A_j(\mathbf{x}) = \nabla_i A_i(\mathbf{x}).$$